A telescope has a magnification of 20 and consists of two converging lenses, the objective and the eyepiece, fixed at either end of a tube 60.0 cm long. Assuming that this telescope would allow an observer to view a lunar crater in focus with a completely relaxed eye, what is the focal length of the eyepiece?

The eyepiece must form its image at infinity.

To find the focal length of the eyepiece, we can use the formula for the magnification of a telescope:

magnification = - (focal length of the objective) / (focal length of the eyepiece)

Given that the magnification is 20 and the focal length of the objective is unknown, we need to solve for the focal length of the eyepiece. Rearranging the formula, we get:

focal length of the eyepiece = - (focal length of the objective) / magnification

Since the eyepiece must form its image at infinity, we can assume that the image formed by the objective is located at the focal point of the eyepiece, which is also at infinity.

Since the observer's eye is completely relaxed, the image formed by the eyepiece will appear to be at infinity. This means that the image formed by the objective is at a distance of 60 cm from the eyepiece.

So, we can substitute the distance between the objective and the eyepiece (60 cm) as the object distance in the lens formula:

(focal length of the objective) = (image distance) * (object distance) / (image distance - object distance)

Since the image is formed at infinity, we can assume the image distance to be very large or approximately equal to infinity. Thus, in the lens formula:

(focal length of the objective) = infinity * (60 cm) / (infinity - 60 cm) ≈ 60 cm

Substituting the values into the formula for the focal length of the eyepiece:

focal length of the eyepiece = - (focal length of the objective) / magnification
focal length of the eyepiece = - (60 cm) / 20
focal length of the eyepiece = -3 cm

Therefore, the focal length of the eyepiece is approximately -3 cm (negative sign indicates that the lens is a diverging lens).